自動控制原理課后答案

ModuIe3

Problem3.1

(a)WhentheinputvariableistheforceF.TheinputvariableFandtheoutputvariableyarerelatedby

theequationobtainedbyequatingthemomentonthestick:

TakingLaplacetransforms,assuminginitialconditionstobezero,

F=-Y+-csY

33

leadingtothetransferfunction

y_31k

Fl+(4c/k)s

wherethetimeconstant「isgivenby

4c

T=——

k

(b)WhenF=0

Theinputvariableisx,thedisplacementofthetoppointoftheupperspring.Theinputvariablexand

theoutputvariableyarerelatedbytheequationobtainedbythemomentonthestick:

k(x—*)l=kl*+Z?c.2l

333dt

TakingLaplacetransforms,assuminginitialconditionstobezero,

3kX=(2k^4cs)Y

leadingtothetransferfunction

Y_3/2

Xl+(2c/A)s

wherethetimeconstant「isgivenby

2c

r

X=Pole.

X

上

2c4c

Problem3.2P54

Detenninetheoutputoftheopen-loopsystem

a

G(s)=-~~-

1+sT

totheinput

r(t)=t

Sketchbothinputandoutputasfunctionsoftime,anddeterminethesteady-stateerrorbetweentheinput

andoutput.ComparetheresultwiththatgivenbyFig3.7.

Solution：

Whiletheinputr(t)=t,useLaplacetransforms,

1

Inputr(s)=—

八/、a1TT

Outputc(s)=r(s)G(s)=—z=a—r1-

S.(1+Ts)sS+1

\T)

thetime-domainresponsebecomes

c(t)=at-aT

Problem3.3

3.3ThemasslessbarshowninFig.P3.3hasbeendisplacedadistancexoandissubjectedtoaunit

impulse3inthedirectionshown.Findtheresponseofthesystemfort>0andsketchtheresultasa

functionoftime.Confirmthesteady-stateresponseusingthefinal-valuetheorem.

Solution：

Theequationobtainedbyequatingtheforce:

kx0+ci0=

TakingLaplacetransforms,assuminginitialconditiontobezero,

KXo+CsXo=l

leadingtothetransferfunction

X_1_11

K+CS~C~^K

c

Thetime-domainresponsebecomes

八1

x(t)=—

Thesteady-stateresponseusingthefinal-valuetheorem:

「「111

limx(/)=lims=—

T8STOK+CssK

K(x0+x)+Ci=方⑺nKx0+X(K+Cs)=1;

1—Kx1—Ka1

?X=-n=

'Cs+KKC

—5-r1

K

/1—KX

x(t)=Q--ec

C

Accordingtothefinal-valuetheorem:

limx(r)=limy-X=lim-~~——=0

T8S->05->0KC.

—5+1

K

Problem3.4

Solution:

l.Iftheinputisaunitstep,then

R(s)=，

s

R(s)、]______C(s)、

->>

\-\-TS

leadingto,

1

C(s)=

(l+rs)s

takingtheinverseLaplacetransformgives,

-t

c(t)=\-er

asthesteady-stateoutputissaidtohavebeenachievedonceitiswithin1%ofthefinalvalue,wecan

solute"t"likethis,

-t

c(t)=\—eT=99%x1(thefinalvalueis1)

hence,

-/

er=°。1(thetimeconstantr=10s)

t-4.605xr=46.05s

2.thenumericalvalueofthenumeratorofthetransferfunctiondoesn'taffecttheanswer.

Seethisequation,

If

_C(s)_A

R(s)1+rs

then

?、A

C(5)=-

(1+7S)S

givingthetime-domainresponse

-i

c(r)=A(l-e7)

asthefinalvalueisA,thesteady-stateoutputisachievedwhen,

-t

C(0=A(1-^7)=AX99%

solutetheequation,

t=4.605T=46.05s

theresultmakenodifferentfromthatabove,sowesaidthatthenumericalvalueofthenumeratorofthe

transferfiinctiondoesn'taffecttheanswer.

Ifa<l,asthetimeincrease,thetwolineswon'tcross.Inthesteadystatetheoutputlagstheinputbyatime

bymorethanthetimeconstantT.

Thesteadyerrorwillbenegativeinfinite.

R(t)

C(t)

t

Ifa=l,asthetimeincrease,thetwolineswillbeparallel.ItisassameasFig3.7.

Ifa>l,asthetimeincrease,thetwolineswillcross.Inthesteadystatetheoutputlagstheinputbyatime

bylessthanthetimeconstantT.

Thesteadyerrorwillbepositiveinfinite.

R(t)

C(t)

t

Problem3.5

Solution:

615+6296145

R(s)=—+-,Y(s)==--++

52?(5s4-1)ss25s+1

.??y(￡)=6f-29+29"〃5

sothesteady-stateerroris29(-30),Toconformtheresult:

limy(t)-lim⑹一29+29,)=8；

Z-?oc—oo

s+6

lim=limy(s)=lim/(s)=lim=00.

fT85—>0ST。S->0s(5s+1)

%=lime(f)=limS?E(S)=limS-[K(5)-R(S)]

iooSTO5->0

=-30

Therefore,thesolutionisbasicallycorrect.

Problem3.6

2夕+3y=x

sinceinputisofconstantamplitudeandvariablefrequency,itcanberepresentedas:

x=4*

asweknow,theoutputshouldbeasinusoidalsignalwiththesamefrequencyoftheinput,itcanalsobe

representedas:

y=

hence

2汝幾e""+3yd“'=4e"0

y。二i

Ao3+2汝

2w

(p=-tan

ItsDC(w—>0)valueis=UJ

Requirement

3

iJAJ=J_X⑷—?w=-

619+402232

.._i71

whilephaselagoftheinput:(p=—tan1=

4

Problem3.7

Onedefinitionofthebandwidthofasystemisthefrequencyrangeoverwhichtheamplitudeoftheoutput

signalisgreaterthan70%oftheinputsignalamplitudewhenasystemissubjectedtoaharmonicinput.

Findarelationshipbetweenthebandwidthandthetimeconstantofafirst-ordersystem.Whatisthephase

angleatthebandwidthfrequency?

Solution：

Fromtheequation3.41

r

A=.°>0.7/;,(1)

Jl+療一

andco>Q(2)

八/,L02

so0<6><

o1.02

sothebandwidthB(o=

fromtheequation3.43

--

thephaseangleZc0=-tanCOT-tan1.02=一

Problem3.8

3.8Solution

AccordingtogeneralizedtransferfunctionofFirst-OrderFeedbackSystems

cKGK

R1+KGH1+K+TS

?thesteadystateoftheoutputofthissystemis2.5V.

.C2.51i

??ifs—>0,>=—.Fromthis,wecangetthevalueofK,thatisK=一.

/?1043

Sinceweknowthatthestepinputis10V,takingLaplacetransforms,theinputis—.

S

Thentheoutputisfollowed

TakingreverseLaplacetransforms,

C=2.5-2.5e%r=2.5(1-e-4,/3r)

Fromthefigure,wecanseethatwhenthetimereached3s,thevalueofoutputis86%ofthesteadystate.So

wecanknow

——=>7=一

l—l"3r=0864占=>7=2

31

Thetransferfunctionis

12+8s4+6s

Let12+8s=0,wecangetthepole,thatisS=-1.5—2/3

Problem3.9Page55

Solution:

Thetransferfunctioncanberepresented,

1+sRC

Leadingtothefinaltransferfunction,

,1+3SRC+(SRC)2

Andthereason:

thesecondsimplelagcompensationnetworkcanberegardedastheloadofthefirstone,andaccordingto

LoadEffect,theloadaffectstheprimaryrelationship;sothetransferfunctionofthecombinationdoesn't

equaltheproductofthetwoindividuallagtransferfunction

ModuIe4

Problem4.1

4.1Theclosed-looptransferfunctionis

10

C—S(S+6)_10

R—l+Tvfc-52+65+10

DIJiVf

Comparingwiththegeneralizedsecond-ordersystem,weget

w?=Vio

2EWn=6

r3V10

10

2

WdJVn>Jl-E=1

Problem4.3

4.3Consideringthespringrisexandthemassrisey.UsingNewton'ssecondlawofmotion

、d(x-y)

my=K(x-y)+c————

dt

TakingLaplacetransfbnns,assumingzeroinitialconditions

mYs2=KX-KY+csX-csY

resultinginthetransferfuncitionwhere

y_CS+K

Xms2+cs+K

And

0=-y/mk

2

c=1.26*105

Problem4.4

Solution:

Theclosed-looptransferfunctionis

￡_1_

J§*S+2=J

RD-Kk1SC2+I2OSC+K

1H?

SS+2

Comparingtheclosed-looptransferfunctionwiththegeneralizedform,

Rs?++①;

itisseenthat

K=(o“

Andthat

2g=2;六

Thepercentageovershootistherefore

P。=100e向=100J

Where

PO<10%

Whensolved,gives1.2<AT(2.86)

WhenKtakesthevalue1.2,thepolesofthesystemaregivenby

1+2s+1.2=0

Whichgives

5=-l±0.45js=-l±1.36j

-0.45

Problem4.5

4.5Aunity-feedbackcontrolsystemhasthefbruzard-pathtransferfunction

S(s+10)

Findtheclosed-looptransferfunction,anddevelopexpressionsforthedampingratio

AnddampednaturalfrequencyintermofKPlottheclosed-looppolesonthecomplex

PlaneforAT=0,10,25,50,lOO.ForeachvalueofKcalculatethecorrespondingdamping

ratioanddampednaturalfrequency.Whatconclusionscanyoudrawfromtheplot?

Solution:

KG(S)

SubstituteG(s)=intothefeedbackformula:①(s)=.Andinunitfeedbacksystem

s(s+10)1+〃G(S)

H=l.

K

Resultin:0(s)=

s2+10.v+^

Sothedampednaturalfrequency,

105

dampingratioQ=--j==.

Thecharacteristicequationiss-+10S+K=0.

WhenK<25,s=-5±,25-K;

WhileK>25,s=—5土ijK-25;

Thevalueof①八and《〔correspondingtoKarelistedasfollows.

K0102550100

-5+V15-5+573/

Pole1S}0-5-5+5i

-5-V15-5-5y/3i

Pole2S2-10-5-5-5i

0Vio55>/210

700后10.5

PlotthecomplexplaneforeachvalueofK:

'im

X-5+5-73:axis

k=100

X-5+5±

k=50

-io-5--s-5+V15

0

XA八）\/real

k=0

k=10k=25k=10k=0axis

Xk=50-5-5i

Xk=i0O-5-

Wecanconcludefromtheplot.

Whenk<25,polesdistributeontherealaxis.ThesmallervalueofKis,thefartherpolesisawayfrompoint

-5.ThelargervalueofKis,thenearerpolesisawayfrompoint-5.

Whenk>25,polesdistributeawayfromtherealaxis.ThesmallervalueofKis,thefurther(nearer)polesis

awayfrompoint-5.ThelargervalueofKis,thenearer(farther)polesisawayfrompoint-5.Andallthe

polesdistributeonalineparallelsimaginaryaxis,intersectrealaxisonthepole-5.

Problem4.6

..v,1fdv.

R+lr+l-FIV,dt+C

RLCRL)bdt

o

TakingLaplacetransforms,assumingzeroinitialconditions,reducesthisequationto

VhRLs

/QLS+7?+RLCs?

Sincetheinputisaconstantcurrentio,so

then,

RL

C（s）=匕

Ls+R+RLCs~

Applyingthefinal-valuetheoremyields

limc(f)=limsC(s)=0

38SfO、J

indicatingthatthesteady-statevoltageacrossthecapacitorCeventuallyreachesthezeroresultinginfull

error.

Problem4.7

4.7Provethatforanunderdampedsecond-ordersystemsubjecttoastepinput,thepercentage

overshootabovethesteady-stateoutputisafunctiononlyofthedampingratio.

Fig.4.7

Solution

Theoutputcanbegivenby

s(§2+

]_________s+2應(yīng)(1)

1(S+血)2+%2(y)

thedampednaturalfrequency①dcanbedefinedas

①d=①“Ji-L

⑵

substitutingaboveresultsin

s+3”______________

22⑶

'S(S+40“>+02(5+^)+<y(/

takingtheinversetransformyields

，一他,

c(f)=l一sm(a)dt+。)

where(4)

themaximumoutputis

叫）=1sin(叫+*

兀兀⑸

I-=-

P%C0n^-^

sothemaximumis

也）=1+產(chǎn)/￡

thepercentageovershootistherefore

PO=100eWG

Problem4.8

Solutionto4.8:

Consideringthemassmdisplacedadistancexfromitsequilibriumposition,

thefree-bodydiagramofthemasswillbeasshownasfollows.

UsingNewton'ssecondlawofmotion,

p—2kx-ex=mx

mx+at+2kx=p

TakingLaplacetransforms,assumingzeroinitialconditions,

X(ms2+cs+2k)=P

resultsinthetransferfunction

X/P=(1/m)/(s2+(c/m)s+2k/m)

=(2/k)(2k/m)(s2+(c/m)s+2k/m)

Aswesee

X(ms2+cs+2k)=P

AsPisconstant

1

SoXoc——.

ms+cs+2k

Whens=——=—6.25x105

2m

(ms2+cs+2k]=105

io4

Xa=k?！?/p>

Thisisasecond-ordertransferfunctionwhere

2

(vn=2k/m

and

-c!2wnm=c/2y12km

Thedampednaturalfrequencyisgivenby

0=①小-U=MklmZ-c218km

=y)2k/m-(c/2m)2

Usingthegivendata,

co,,=V2X5X104/2X106=V^05=0.2236

250

=2.7950x1O-4

-2X2X106XV^05

織=0.2236xjl-(2.7950x1()-4)2=0.2236

Withthesedatawecandrawapicture

71

T,=—=14.050

3d

-=16000

3

Ts=4.6rf=73600

X112k/in1%

Pms?+cs+2k2k^,2+5+mcod(s+血)+G；

mm

甘山恢丁康片cc

其中叫二匕.而‘叫氣盛"兩‘血=不

-￡-儂"

/.x=?Psina)dt

mcod

???K=,,=篇e「口'"(—3"sincodtp+%cos*)=0

tan①/p-~~=>*)=7.03=>xp-0.02m

Problem4.10

4.10

solution:

ThesystemissimilartotheoneinthebookonPAGE58toPAGE63.Thedifferenceistheconnectionof

thespring.Sothetransferfunctionis

a=

a$2+27%普+/2

仇=k&k“,N

2

ed_RJs+RCs+(R+kakmN)kp

22

J=NJm+J?C=Ncm+cl-,

N43旦

N,“助仇

_/K?K,

W"~vNJ'R

dampingratio

ButthevalueofJisdifferent,becausethereisaspringconnected.

"N2N2

Becauseoffinal-valuetheorem,

a

a27

ModuIe5

Problem5.4

5.4Theclosed-looptransferfunctionofthesystemmaybewrittenas

10K

c=S2+6S+10=_K_10(K+l)

R]I10K1+K52+6S+10A：+10

S2+65+10

Theclosed-looppolesarethesolutionsofthecharacteristicequation

旦旺叵一土后加j

2

叱,=J10(l+K)

2EjlO(l+K)=6

,10(1+K)

Inordertostudythestabilityofthesystem,thebehavioroftheclosed-looppoleswhenthegainK

increasesfromzerotoinfintewillbeobserved.Sowhen

&=2E=—S=-3土后J

'10

K,=10E=S=-3±V10TJ

2110

K、=20E=^^-S=-3±V201J

370

雙擊下面可以看到原圖

Re

Problem5.5

Solution

Theclosed-looptransferfunctionis

K

Ri+3(")S2+K(1+〃S)d+oKs+K

s2

Comparingtheclosed-looptransferfunctionwiththegeneralizedform,

C=成

Leadingto

CDn=y[K

ea尿

Thepercentageovershootistherefore

Y冗

PO=100e貨=40%

Producingtheresult

1=0.869(0.28)

Andthepeaktime

Leadingto

%=1.586(0.82)

Problem5.7

5.7ProvethattherisetimeTrofasecond-ordersystemwithaunitstepinputisgivenby

1CD(i1-Jl二42

「=￡tan」嬴=-tan1'—~一

Plottheriseagainstthedampingratio.

Solution:

Accordingto(4.33):c(t)=l-e(cos(odt+."sincodt).4.33

Whent=T.,c(t)=l.substituec(t)=1into(4.33)

Producingtheresult

1(D(l1-Jl-42

Tr=mtan」嬴=晟tan-1-[-

Plottherisetimeagainstthedampingratio:

Tr!s

Problem5.9

Solutionto5.9:

Asweknowthatthesystemistheopen-looptransferfunctionofaunity-feedbackcontrolsystem.

SoGH(S)=G(S)

Givenas

GH。)；^~~-

'"2)(S+5)

Theclose-looptransferfunctionofthesystemmaybewrittenas

As)_G(s)_4K

R''l+GH(s)(s-2)(s+5)+4K

Thecharacteristicequationis

(s—2)(s+5)+4K=0ns2+3s+4K—10=0

AccordingtotheRouth'smethod,theRouth'sarraymustbeformedasfollow

5214K-10

s30

5°4K-10

Forthereisnoclosed-looppolestotherightoftheimaginaryaxis

4K-1020nKN2.5

Giventhat=0.5

CDn—J4K-10

3

0=-.，nK=4.75

2J4K-10

WhenK=0,therootare

s=+2,?5

Accordingtothecharacteristicequation,thesolutionsare

whileK<3.0625,wehaveoneortwosolutions,allareintegralnumber.

Orwewillhavesolutionswithimaginarynumber.

Sowecandraw

□Open-looppoles

Closed-looppoles

Problems.10

5.10

solution:

<=0.6

w=2rad/s

accordingto

e-Cw-'1

(c)=l/sin(叼+。)=7

2

e-12,?sin(l.6/+^)=0.4tan0=g

finally,tisdelaytime:

ta1.235(0.67)

ModuIe6

Problem6.3

FirstweassumethedisturbanceDtobezero:

e=R-C

C^K-e---

5+1S

e_s(s+l)

Hence:

元-10K+s(s+l)

ThenwesettheinputRtobezero:

C=(Ke+D)?——=-en—=-

s(s+l)D10K+s(s+l)

Addingthesetworesultstogether:

s(s+l)n10n

10K+s(s+l)10K+s(s+l)

R（S）W；*

_5+110_5-9

e―To&+7(7+i)―io^7+7(7+i)―ioo5+?(5+i)

thesteady-stateerror:

”—9vv—9

=lims?e=lim--4——=lim=-0.09

STOI。/+$2+]oosiOS+S+100

Problem6.4

Detenninethedisturbancerejectionratio(DRR)forthesystemshowninFigP.6.4

+

fig.P.6.4

solution：

fromthediagramwecanknow:

&=0.2

R

(=1

c=0.05

sowecangetthat

==1+￡A_=1+22rl=5

SgJcLcR0.05

2],soc=0.025,DRR=9

0.15+0.050.05s+0.025

Problem6.5

6.5Solution

Forthepurposesofdeterminingthesteady-stateerrorofthesystem,weshouldgettoknowtheeffect

oftheinputandthedisturbancealongwhentheotherwillbeassumedtobezero.

Firsttosimplifytheblockdiagramtothefollowingpatter:

Allowingthetransferfunctionfromtheinputtotheoutputpositiontobewrittenas

■=20

2

edJS+2S+20

e200:20*2一40

01-222

Js+2.V+20廣js+2s+20s-(Js+2s+20)$

AccordingtotheequationE=R-C:

e")=解網(wǎng)⑸—織切=翹儂—不言而)］寸巴曲卷

0.2

問題；

1.系統(tǒng)型為2,對于階躍輸入，穩(wěn)態(tài)誤差為0.

2.終值定理寫的不對。

andthetransferfunctionfromtheinputtotheoutputpositiontobewrittenas

1

TdJs?+2s+20

Q-___________T二_______________

02J?+25+20"S(J$2+2S+20)

Accordingtotheequatione="c

1-1

%，⑸=媽［一%⑸］=媽［一百+25+201=/+2S+20-0.05

s。thestateerrorshouldbe:%=+e對=0.2-0.05=0.15

Problem6.9

Solution:

Thetransferfunctionoftheinsideloopis

10

fJ-5-(-5---+--1--)---1-0

1+lOZs-$2+(10&+1)5

s(s+1)

Andthetransferfunctionofthewholesystemis

C__________10

J-52+(10^+1)5+10

Wegetthevalueof叼and，fromtheaboveequation:

wn=Vio

10)+1

_2>/10

AndthevalueofPOis

P0=100e-s/席

Accordingtotheformerequation:

PO=10%n?=0.6=

2V10

Thefinalresultis

kJ2布-1=028

10

ModuIe7

Problem7.5

Determinetheclosed-looptransferfunctionandthepercentageovershootfbrasystemdescribedby

thepole-zeromapshowninFig.P7.5,assumingthesteadystategainoftheclosed-looptransferfunctionis

unity.

頓1

(a)

-i

SOLUTION：

Fromthefigure,wecangettheBodeformfunction

C_1+sr

Rs~/co~+2s/co

。n+1

]_電、=J(_2/+F=#)—

3

1/r_3

，=0.894a0.9

菽=2

FromFig.7.6onpage117,wecanseethat,undertheeffectofclosed-loopzero,while0=0.9and

/=—,thePercentovershootislessthan1%.

2

C1

R-(1+57)。2/d+2公/%+1)

—l/r=-3=>r=l/3

3'=2,?y?=V5?2.24n4=2/石。0.894,

C15

R(s+3)(52+4s+5)

Problem7.7

7.7Asystemhasatransferfunctionthatmaybewritteninthefonn

C_1

R(s+l)($2+〃s+b)

Itisknownthatforthesecond-orderterm,。=0.2.Investigationoftheunitstepresponserevealsan

overshootof5%oftheinput.Calculatetheconstantsaandb,plottheclosed-loopsystempolesonthe

complexplane,andcommentonthereducabilityofthesystem.

Solution:

Thethird-ordersystemwith(=0.2wouldhaveanovershootof5%iftheadditionalpolewerelocated

suchthattheparameter(3=2.25.

then

￡=4==2.25(看圖似不超過2.1)

初0.2例

Thetime-domainresponseparametersmaybecalculateas

(on=4rad/sand7=0.2

a=23口=1.6andb=*=16

thenwewillplotclosed-loopsystempolesonthecomplexplane:

Theotherrealpolecannotbeneglected,theeffectofthereal-axispolecontributiontotheresponsewillbe

tomakeitmoresluggish.

Problem7.9

Solution:

Examinationoftheplantrevealsthefollowingtime-domainparameters:

co=2rad/sJ=1—=/?;

nT

ExaminingFig7.6,thesystemwithJ=1wouldhaveanovershootof5%ifthezeroislocatedsuchthat

theparameter/=0.6.Thisrelatedtotheparameterbby

y2

y=0.6==一

她b

b=\.2

Iftheinputisaunitstep,thesteady-stateout